Guys please, please help!! 1. What is the domain and range?

Mathematics

1 answer:

Vilka [71]3 years ago

7 0

Answer:

<h2>Domain is a ser of possible input values</h2>

<h3>Range:is a set of Y coordinates</h3>

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Help me pls due tmrw

stiv31 [10]

( a )pairs of vertical angles :

AGB and EGD

Vertical angles are a pair of opposite angles formed by intersecting lines. Vertical angles are equal in magnitude.

( b ) a pair of adjacent angles :

AGB and BGC

Adjacent angles are two angles that have a common side and a common vertex (corner point) but do not overlap in any way.

AGB and BGD

Two angles are called supplementary when their measures add up to 180 degrees.

(d) A pair of complementary angles :

AGF and FGE

Two angles are called complementary when their measures add to 90 degrees.

(e) Find measure of CGD :

( look into the photo attached for better understanding)

CGD = 55°

<em>hope this helps :)</em>

8 0

2 years ago

Both circle Q and circle R have a central angle measuring 140°. The area of circle Q's sector is 25π m2, and the area of circle

taurus [48]

Solution:

we are given that

Both circle Q and circle R have a central angle measuring 140°. The area of circle Q's sector is 25π m^2, and the area of circle R's sector is 49π m^2.

we have been asked to find the ratio of the radius of circle Q to the radius of circle R?

As we know that

Area of the sector is directly proportional to square of radius. So we can write

$Area \ \ \alpha \ \ r^2\\ \\ \Rightarrow A=kr^2\\ \\ \text{where k is some proportionality constant }\\ \\ \Rightarrow \frac{A_1}{A_2}=\frac{r_1^2}{r_2^2}\\ \\ \Rightarrow \frac{r_1^2}{r_2^2}=\frac{A_1}{A_2}\\ \\ \Rightarrow \frac{r_1^2}{r_2^2}=\frac{25 \pi}{49 \pi}\\ \\ \Rightarrow \frac{r_1}{r_2}=\sqrt{\frac{25 \pi}{49 \pi}} \\ \\ \Rightarrow \frac{r_1}{r_2}=\frac{5}{7} \\$